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Suppose $V$ is a vector space over a scalar field $F$. If $\dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?

My thought was that if $\mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $\mathscr{B}\to F$, by identifying elements of $V$ with their $\mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.

Is this correct? If so, how does one find the cardinality of $\{f\colon\mathscr{B}\to F\mid \mathrm{supp }(f)<\infty\}$, in terms of say $|F|$ and $|\mathscr{B}|$? Thanks.

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Can you find the number of functions from $\mathscr{B}$ to $F$ with support of size at most $n$? – Chris Eagle Sep 11 '12 at 18:48
@ChrisEagle Wouldn't that require choosing $n$ vectors in $\mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $\mathscr{B}$ is infinite. – Nastassja Sep 11 '12 at 19:10
@Nastassja But what would the infinite cardinal be? – Alex Becker Sep 11 '12 at 19:10
@Nastassja The point is that the set of functions from $\mathscr B$ to $F$ with support at most $n$ is a union of at most $|\mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic. – Alex Becker Sep 11 '12 at 19:19
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=\max\{|B|,|F|^n\}=\max\{|B|,|F|\}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $\max\{|B|,|F|\}\cdot\aleph_0=\max\{|B|,|F|\}$ anyway since $|B|\geq\aleph_0$? – Nastassja Sep 11 '12 at 19:28
up vote 6 down vote accepted

Suppose that $V$ is a vector space over $F$ and $V$ has a basis $B$.

From the definition of a basis every $v\in V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $B\times (F\setminus\{0\})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.

This gives a well-defined injection from $V$ into finite subsets of $B\times(F\setminus\{0\})$. Assuming the axiom of choice we have that, $$|V|\leq\left|[B\times(F\setminus\{0\})]^{<\omega}\right|=|B\times F|=\max\{|B|,|F|\}\leq|V|\implies|V|=\max\{|B|,|F|\}.$$

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The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field. – Chris Eagle Sep 11 '12 at 19:06
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space. – Nastassja Sep 11 '12 at 19:17
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-) – Asaf Karagila Sep 11 '12 at 20:02
Thanks Asaf, I think this is a much neater presentation than what I said above. – Nastassja Sep 11 '12 at 21:10
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-) – Asaf Karagila Sep 11 '12 at 21:12

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